1. Define the areas:
- Let $Z$ be the area commonly bounded by the triangle and the circle.
- Let $X$ be the area of the shaded region.
- Let $Y$ be the area of the starred region.

2. Set up the equations:
- The area of the circle is composed of $Z$ and three times the area of the starred region $Y$:
$$\text{Area of the circle} = Z + 3Y$$
- The area of the triangle is composed of $Z$ and three times the area of the shaded region $X$:
$$\text{Area of the triangle} = Z + 3X$$

3. Subtract the two equations:
$$\text{Area of the triangle} - \text{Area of the circle} = (Z + 3X) - (Z + 3Y)$$
Simplifying, we get:
$$\text{Area of the triangle} - \text{Area of the circle} = 3X - 3Y$$
Therefore:
$$3X - 3Y = \text{Area of the triangle} - \text{Area of the circle}$$

4. Calculate the area of the equilateral triangle:
- The formula for the area of an equilateral triangle with side length $s$ is:
$$\text{Area of the triangle} = \frac{\sqrt{3}}{4} s^2$$
- Given $s = 5$:
$$\text{Area of the triangle} = \frac{\sqrt{3}}{4} \cdot 5^2 = \frac{25\sqrt{3}}{4}$$

5. Calculate the area of the circle:
- The formula for the area of a circle with radius $r$ is:
$$\text{Area of the circle} = \pi r^2$$
- Given $r = 2$:
$$\text{Area of the circle} = \pi \cdot 2^2 = 4\pi$$

6. Substitute the areas into the equation:
$$3X - 3Y = \frac{25\sqrt{3}}{4} - 4\pi$$

7. **Solve for $X - Y$:**
$$X - Y = \frac{3X - 3Y}{3} = \frac{\frac{25\sqrt{3}}{4} - 4\pi}{3} = \frac{25\sqrt{3} - 16\pi}{12}$$